3

在以下真实世界示例中，我进行了匹配：

type Style = Nice | Cool | Ugly
type Color = Blue | Yellow | Orange | Grey | Cyan
type ClothingProperties = Style * Color

type Clothes =
| Jeans of ClothingProperties
| Pullover of ClothingProperties
| Shirt of ClothingProperties

type Person =
| Person of string * Clothes

let team = [Person("Jan", Jeans (Cool, Blue)); Person("Pete", Shirt (Nice, Cyan)); Person("Harry", Pullover (Ugly, Grey))]

let matchPerson person=
    match person with
    | Person(name,  Jeans(Ugly,_) ) -> printfn "%s wears ugly stuff." name
    | Person(name,  Pullover(Ugly,_) ) -> printfn "%s wears ugly stuff." name
    | Person(name,  Shirt(Ugly,_) ) -> printfn "%s wears ugly stuff." name
    | _ -> ()

List.iter(fun x->matchPerson x) team

有没有办法创造更有效的搭配，这样我就不需要检查每个衣箱了？像这样的东西：

let matchPerson person=
    match person with
    | Person(name,  _ (Ugly,_) ) -> printfn "%s wears ugly stuff." name
    | _ -> ()

当然，这不是正确的语法。但是我怎样才能达到这样的效果呢？

4

1 回答 1

5

这并不简单，您可以使用反射，但问题是您的可区分联合需要重新设计，因为如果您知道总会有 ClothingProperties，那么您可以将其更改为：

type Style = Nice | Cool | Ugly
type Color = Blue | Yellow | Orange | Grey | Cyan
type ClothingProperties = Style * Color // or just use a tuple

type Clothe =
| Jeans 
| Pullover
| Shirt

type Clothes = Clothe *ClothingProperties
type Person =
| Person of string * Clothes

let matchPerson person=
    match person with
    | Person(name,  (_,(Ugly,_)) ) -> printfn "%s wears ugly stuff." name
    | _ -> ()

此处描述了一个相关问题是否可以将有区别的联合标签作为参数传递？

于 2014-04-11T09:56:31.160 回答